= Solution
Here the star means entrywise <complex conjugation>, not <Hermitian conjugation>. Put $s=g_3$ and $r=g_1g_3$. Since $\alpha^3=1$ and $1+\alpha+\alpha^2=0$,
$$
r=\begin{pmatrix}\alpha^2&0\\0&\alpha\end{pmatrix},\qquad
s=\begin{pmatrix}0&1\\1&0\end{pmatrix},\qquad
r^3=s^2=I,\qquad srs=r^{-1}.
$$
The generators are $g_1=rs$, $g_2=r^2s$ and $g_3=s$. Moving every occurrence of $s$ to the right using $sr=r^{-1}s$ reduces every word to $r^k$ or $r^ks$ with $k=0,1,2$. The three diagonal <matrices> are distinct, as are the three off-diagonal <matrices>, so
$$
\boxed{G=\{I,r,r^2,s,rs,r^2s\},\qquad |G|=6.}
$$
For an explicit <group isomorphism>, send $r$ to $(123)$ and $s$ to $(12)$ in the <symmetric group> $S_3$, composing <permutations> from right to left. These <permutations> satisfy the same relations and generate $S_3$. The normal forms above then give an isomorphism; in particular $g_1\mapsto(13)$, $g_2\mapsto(23)$ and $g_3\mapsto(12)$.
Conjugation by $s$ exchanges $r$ with $r^2$, and conjugation by $r$ cycles the three reflections. The <conjugacy classes> are therefore
$$
\boxed{\{I\},\qquad \{r,r^2\},\qquad \{s,rs,r^2s\}=\{g_1,g_2,g_3\}.}
$$
Their sizes are $1,2,3$, respectively. They correspond to the identity, three-cycles and <transpositions> in $S_3$.
Back to article page
